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Originally posted by Nordlysthe C is short for combination, and the P in his first post referred to permutation.
I am not sure what the "6 C 2" means, but yes, 15 is right. Every player has to play five games, so that's 30 games divided by 2 because there are always two players involved.
combination is where the order doesn't matter. e.g. {abc}={bca}={cba}..., while in a combination the order does matter, {abc} does not equal {bca}.
the formulas are,
n P r=P(n,r)=n!/[(n-r)!]
n C r=C(n,r)=n!/[r!(n-r)!]
in the example here, the order does not matter. 1A=A1.
8 C 8=8!/0!=8!
Originally posted by RahimKJust 1
Just to refresh my memory:
If you have 8 proffessors and 8 offices and you have to assign each to 1 office how many different ways are there?
8 P 8 = 40, 320 or 8! = 40, 320.
This is it right?
Just a sidenote here which I was thinking about. You have a chess tourney and you have 6 players and it's a round robin. How many total games?
6 C 2 = 15 right?
Because you are in charge and get to make the selection!
Harri / Luck
Originally posted by wormwoodThey will show up during their scheduled office hours.
zero, as none of the professors ever show up in their offices...
The easiest (and just about the only way) way to remove permanent academic staff from the university is to complain that they are not in the office during their scheduled hours.
They may not spend a lot of time of the campus, but tehy will respect the stated office hours or they will get fired!
Harri / Luck